Question

A circular current carrying coil has a radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be ${\frac{1}{8}}^{th}$ to its value at the centre of the coil, is :

• A. $\frac{R}{\sqrt{3}}$
• B. $R\sqrt{3}$
• C. $2\sqrt{3}R$
• D. $\frac{2}{\sqrt{3}}R$

Answer 1

Answer: (B)

$R\sqrt{3}$

The magnetic field at the center of a current carrying circular loop of radius R is:
$B_{center}=\frac{{\mu }_{0}i}{2R}$
The magnetic field on the axis at a distance x from the center of a
current carrying circular loop of radius R is:
$B_{axis}=\frac{{\mu }_{0}i{R}^2}{{(R^2+x^2)}^{\frac{3}{2}}}$
Given,
$B_{axis}=\frac{1}{8}{B}_{center}$
$\therefore 8\frac{{\mu }_{0}i}{2R}=\frac{{\mu }_{0}i{R}^2}{(R^2+x^2{)}^{\frac{3}{2}}}$
$\Rightarrow \frac{{(R^2+x^2)}^3}{R^6}=64$$\Rightarrow {\left(\frac{R^2+x^2}{R^2}\right)}^3=4^3$
$\Rightarrow \frac{R^2+x^2}{R^2}=4$$\Rightarrow 1+\frac{x^2}{R^2}=4$
$\frac{x^2}{R^2}=3$
$x=\sqrt{3}R$